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I have some questions regarding the Ising model in the presence of a magnetic field which is non-uniform.

Let us define the Ising Hamiltonian on a $d-$dimensional lattice,

$$ H = -\frac{1}{2} \sum_{i,j}J \sigma_i \sigma_j - \sum_i h_i \sigma_i,$$

where $J$ is the interaction parameter, $h_i$ the magnetic field on site $i$, and $\sigma_{i,j}$ the spin variable on sites $(i,j)$ ($i.e.$, $\sigma_{i,j} = +1$ or $-1$).

I have two questions:

(1) When $h$ is uniform, according to the Lee-Yang theorem, there is no phase transition in the thermodynamic limit with respect to temperature (I guess, $\forall J$ positive or negative). Does this result apply for any dimension $d$ and in particular, in the infinite dimensional lattice (= mean-field) case? Also, does it depend on the lattice structure ?

(2) When $h$ is non-uniform, is there any phase transition with respect to temperature ? In particular, I am interested in the mean-field case. I found some solutions in this paper. It said that "for any positive (resp. negative) and bounded magnetic field, the model does not present the phase transition phenomenon whenever lim inf $h_i > 0$", but I am not sure to understand. Does that mean that if the minimum value of the magnetic field over all the lattice is positive and not infinite, then there is no phase transition ?

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  • $\begingroup$ 1) yes, the field forces all spins to align in some direction at all temperature. Therefore, there is no symetry to break. $\endgroup$
    – Syrocco
    Commented Jun 6 at 7:19
  • $\begingroup$ 2) see for example models with a random field (look at random ising field model) where phase transition are possible above d=2 $\endgroup$
    – Syrocco
    Commented Jun 6 at 7:19

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First, considering the Lee-Yang theorem in its usual form (i.e., with constant magnetic field $h$ and 2-body interactions):

  • The theorem holds only for ferromagnetic interactions. The result is actually false in general if one removes this assumption (see, for instance, Example 8.17 in this paper).
  • It holds on any graph (provided that the interactions decay fast enough for the thermodynamic limit to be well defined). The proof is usually done first for finite systems, and in this case the structure of the graph is completely irrelevant.
  • The result of the circle theorem remains true in the mean-field case (i.e., for the Curie-Weiss model). In fact, much more can be said about the asymptotic density of the zeroes on the unit circle; see this paper.

For your question on the case of inhomogeneous magnetic field, it is actually proved in the paper you mention that there is no phase transition when the magnetic field is positive at each site and $\liminf_{i\in\mathbb{Z}^d} h_i > 0$; see Theorem 4 in the paper.

Finally, there exist various extensions of the Lee-Yang theorem. I refer to this review paper for more information.

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  • $\begingroup$ Perfect, thank you ! Regarding the question on the inhomogeneous case, I am not fully familiar with limit notations: by lim inf$_{i \in Z^d} h_i >0$, do we mean that the smallest value of the sequence $h ={h_1, h_2, ..., h_N}$ where $N$ is the total number of lattice sites, should be positive? (So I need to check on all the lattice sites the smallest value and this value should be positive). Also, do we agree if we say that in the paper, the theorem 4 does not work if at one lattice point, the field diverges? So the result is only OK for a bounded magnetic field...? $\endgroup$
    – math-int
    Commented Jun 6 at 12:18
  • $\begingroup$ I understand their notation as saying that $\lim_{R\to\infty}\inf_{i:\|i\|>R} h_i > 0$. And yes, they seemingly need bounded fields, although I don't know whether this is purely technical and I don't really have time to think about that now. $\endgroup$
    – Yvan Velenik
    Commented Jun 6 at 13:10
  • $\begingroup$ Thank you, all good ! Again, by "lim𝑅→∞inf_{𝑖:‖𝑖‖>𝑅} ℎ_𝑖>0", can you translate that into an english sentence please ? $\endgroup$
    – math-int
    Commented Jun 6 at 15:19
  • $\begingroup$ One can reformulate the condition as follows: there exist $R<\infty$ and $c>0$ such that, for all $i$ such that $\|i\|\geq R$, one has $h_i\geq c$. $\endgroup$
    – Yvan Velenik
    Commented Jun 6 at 16:32

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